What is the integral of 8(tan^2(x)+tan^4(x)) ?

The integral of 8(tan^2(x)+tan^4(x)) is 8/3 tan^3(x)+C

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integral of 8(tan^2(x)+tan^4(x))

\int 8 (tan^{2} (x) + tan^{4} (x)) d x

\frac{8}{3} tan^{3} (x) + C

Solution steps

\int 8 (tan^{2} (x) + tan^{4} (x)) d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 8 \cdot \int tan^{2} (x) + tan^{4} (x) d x

Rewrite using trig identities

= 8 \cdot \int - 1 + sec^{2} (x) + tan^{4} (x) d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 8 (- \int 1 d x + \int sec^{2} (x) d x + \int tan^{4} (x) d x)

\int 1 d x = x

\int sec^{2} (x) d x = tan (x)

\int tan^{4} (x) d x = \frac{1}{3} tan^{3} (x) + x - tan (x)

= 8 (- x + tan (x) + \frac{1}{3} tan^{3} (x) + x - tan (x))

Simplify

8 (- x + tan (x) + \frac{1}{3} tan^{3} (x) + x - tan (x)) : \frac{8}{3} tan^{3} (x)

= \frac{8}{3} tan^{3} (x)

Add a constant to the solution

= \frac{8}{3} tan^{3} (x) + C

x, \frac{1}{2} x, 9

x \to - 1 + lim (2 x^{3} - 2 m)

f (x) = 5 e^{2 x}

\frac{d y}{d x} = \frac{e ^{x}}{8 + e ^{x}}

\frac{d}{d x} (e^{6 x^{2}})

What is the integral of 8(tan^2(x)+tan^4(x)) ?
The integral of 8(tan^2(x)+tan^4(x)) is 8/3 tan^3(x)+C